VCE Specialist 3/4 Question Thread!

[Stevensmay]

The angle formed between the line and the x/y axes is the same, pi/4 in both cases. There is nothing else that I can think of that would make them different.
Unless some weird restriction would affect one but not the other, not exactly sure if that is doable.

[SocialRhubarb]

Let z=x+yi

[Alwin]

Hey there, I'm just wondering whether the following diagram could also be expressed as Re(z) + Im(z) as well? (I know C is incorrect).

It's from the 2001 VCAA exam by the way, cheers.

You can also do it graphically

Remember that for |z-z₁| = |z-z₂|, geometrically it represents the line that perpendicularly bisects the line between  z₁ and z₂
That is why in the diagram they draw the line between (2, 0) and (0, -2)

more of my... amazing *cough cough* paint skills


hope it helps + makes sense

[saba.ay]

Could someone please explain how to do part b?

Please and thanks

[lzxnl]

Part a: pretend that y=0 at where the baseball is hit. Max height = 6.5m
The x component of the velocity is irrelevant here, so focus on the y component only.
v^2 = u^2 + 2ay
At the top of the motion, v=0. a=-g
u^2 -2gy=0 y=6.5
u^2-13g=0.
Find u. This will be the y component of the initial velocity. Work out the x component from this and the given angle.

Part b:
Now you want the time it takes for the ball to hit the ground. In the y direction, y=ut-1/2 gt^2
You know what u is and that y=-1.5 (ball's height drops by 1.5m). Find t.
Then, multiply this time by the x component of the velocity as the horizontal acceleration is zero to find the horizontal displacement. You have horizontal and vertical displacements. Work out distance from this.

Part c:
x component of the velocity is constant. For the y component, use v^2 = u^2 - 2gy again. To avoid any errors from part a, let the u be at the top, where u=0. Hence y in this case is a drop of 8 metres, so -8. v^2=16g. Solve and find the vertical velocity. You have the horizontal and vertical velocities. Find the angle and speed.

[saba.ay]

Part a: pretend that y=0 at where the baseball is hit. Max height = 6.5m
The x component of the velocity is irrelevant here, so focus on the y component only.
v^2 = u^2 + 2ay
At the top of the motion, v=0. a=-g
u^2 -2gy=0 y=6.5
u^2-13g=0.
Find u. This will be the y component of the initial velocity. Work out the x component from this and the given angle.

Part b:
Now you want the time it takes for the ball to hit the ground. In the y direction, y=ut-1/2 gt^2
You know what u is and that y=-1.5 (ball's height drops by 1.5m). Find t.
Then, multiply this time by the x component of the velocity as the horizontal acceleration is zero to find the horizontal displacement. You have horizontal and vertical displacements. Work out distance from this.

Part c:
x component of the velocity is constant. For the y component, use v^2 = u^2 - 2gy again. To avoid any errors from part a, let the u be at the top, where u=0. Hence y in this case is a drop of 8 metres, so -8. v^2=16g. Solve and find the vertical velocity. You have the horizontal and vertical velocities. Find the angle and speed.

Thank you. Couldn't seem to make the connection that y=-1.5 -.- Much appreciated

[Jeggz]

This is probably a really basic question, but I've absolutely forgotten everything to do with vectors :\
Anyways, how do you find the magnitude of a vector when it's not in its i,j,k form? You only know it in its a,b form for instance?

[Stevensmay]

What do you mean by a and b sorry? Co ordinates?

[Jeggz]

Sorry! Haha, I knew I didn't phrase that properly!
As in, you know the vector AB = b-a, or the vector AC = c-a ? (with tildas on the bottom)

[lzxnl]

Sorry! Haha, I knew I didn't phrase that properly!
As in, you know the vector AB = b-a, or the vector AC = c-a ? (with tildas on the bottom)

|a-b|^2 = (a-b).(a-b) = |a|^2 + |b|^2 - 2a.b
so square root both sides

[Jaswinder]

A body of mass 5kg is on a rough horizontal table, the coefficient of friction being u. It is connected by a horizontal light inextensible string, which passes over a smooth peg, to a body of mass 3kg which hangs vertically.
If the end of the table containing the peg is lowered so that the table makes a downward angle of 30 degrees with the horizontal, find acceleration:

i get the right force but by what masss should i divide by 5? or 8?

b) find the tension in the string

don't know how to?

[lzxnl]

A body of mass 5kg is on a rough horizontal table, the coefficient of friction being u. It is connected by a horizontal light inextensible string, which passes over a smooth peg, to a body of mass 3kg which hangs vertically.
If the end of the table containing the peg is lowered so that the table makes a downward angle of 30 degrees with the horizontal, find acceleration:

i get the right force but by what masss should i divide by 5? or 8?

b) find the tension in the string

don't know how to?

Which body are you analysing? Personally, I would look at it this way.
You can treat the whole system as one block of 8 kilograms. The downwards forces on it are the weight force (components), given by 3g+5g sin 30=5.5g
The upwards force on it is the friction force, given by 5*u*cos30
Subtract these to get 5.5g - 5*u*sqrt3 /2 = 8a. Divide by 8 as we're considering all of the forces in the 8 kilogram system.
You have your acceleration.

OR: You could do it this way.
Forces on the 3kg object: moving downwards, so treating downwards as positive
3g-T=3a
Forces on the 5kg object:
5g*sin 30 + T - u5g cos 30 = 5a
Adding to eliminate the T's, and we have the equation given above.

How to find the tension? Look at the forces on the 3 kg block. 3g - T = 3a. You have a => find T.

[sin0001]

Is there a way to 'visualise' and deduce the direction of Tension experienced by a string/rope? I don't have a specific question to clarify what I mean, but I was wondering if there was a general rule to figure this out.
Thanks

[lzxnl]

What do you mean? Tension acts always to pull the object it is connected to in the direction of the rope.
As an example, if you have a block and a rope is connected to the left of the block, any tension that exists will attempt to pull the block to the left.

[Alwin]

Is there a way to 'visualise' and deduce the direction of Tension experienced by a string/rope? I don't have a specific question to clarify what I mean, but I was wondering if there was a general rule to figure this out.
Thanks

Tension... always acts along the rope/string/cord/vine/whatever vcaa wants Tasmanian Jones to swing across with

So its just 2 arrows acting on the two different objects and they point towards each other (of same magnitude) along the rope.

EG:
Rope~      ============================ 
Tension~  ------>                                         <--------

Hope it helps and I didn't just make a fool of myself


EDIT: beaten